An archer applies an averafge force of 200 N to draw the bow string back 1.3m.

How much mechanical potential energy is stored in the streched bow?

How much kinetic enery does the arrow have before it is released?

How much potiential energy will the arrow have after it leaves the bow string?

If a 0.3 kg arrow is shot from this bow, then how fast will it be moving just after it leaves the bow string?

The concepts that you need to do these problems are:

A. The work you do on an object is equal to the force you apply times the distance you move it:

W = F x D.

B. When you pull back on the draw string that work is momentarily stored as potential energy in the bow (until you release the arrow):

PE = W.

C. When you release the arrow all the potetial energy that was stored in the bow is converted to the kinetic energy of the arrow.

D. The kinetic energy of an object is equal to:
KE = 1/2 mv^2


You can solve the first question using principles A and B, the second and third question using C, and the last question using D. So now please try and work these problems through, and post back with your answers.

OK FOR THE FIRST QUESTION.

W= F * D
200 N * 1.3 M
260 j

PE= W
PE= 260 J

IS THIS CORRECT FOR THE FIRST ANSWER?

OK FOR THE FIRST QUESTION.

W= F * D
200 N * 1.3 M
260 j

PE= W
PE= 260 J

IS THIS CORRECT FOR THE FIRST ANSWER?

Yes - you're doing great!

OK. What is the mass and velocity in the equation ke=1/2mv^2

mass = mass of the arrow = 0.3 Kg. Velocity of the arow is what you need to calculate, based on the KE of the arrow. You know the KE of the arrow as it leaves the bow is equal to the amount of PE that you put into the bow.

I got v= 41.6 m/s

i got v= 41.6 m/s
Right!

OK is this correct for question b?

How much kinetic energy does the arrow have before it is released?

ke=1/2mv^2
=1/2(0.3)(41.6)^2
=259.6

No - you used the velocity of the arrow after it is released - but the question asks about the KE of the arrow before it is released. So think about that - when the arrow is pulled back, ready to be released (but not released yet), what is its velocity? And from that, what is its KE at that instant?

I said the velocity is 0

So is the velocity 0 at that point

So is the velocity 0 at that point

Yes. So its KE at that point is?

The ke at that point is 0 as well